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3 years ago

2. Two cars A and B are heading forward. The velocity of A and B are 30 m/s and 20 m/s

Physics

1 answer:

KengaRu [80]3 years ago

8 0

Answer:

A.) 27000 kgm/s

18000 kgm/s

B.) Va = 22 m/s

C.) 19800 kgm/s

25200 kgm/s

Explanation: Given that the velocity of A and B are 30 m/s and 20 m/s. And of the same mass M = 9 × 10^5g

M = 9×10^5/1000 = 900 kg

A.) Initial momentum of A

Mu = 900 × 30 = 27000 kgm/s

Initial momentum of B

Mu = 900 × 20 = 18000 kgm/s

B.) if they have an accident and then the velocity of the B is 28 m/s, find out velocity of A.

Momentum before impact = momentum after impact

Given that Vb = 28 m/s

27000 + 18000 = 900Va + 900 × 28

45000 = 900Va + 25200

900Va = 45000 - 25200

900Va = 19800

Va = 19800/900

Va = 22 m/s

C.) Momentum of A after impact

MV = 900 × 22 = 19800 kgm/s

Momentum of B after impact

MV = 900 × 28 = 25200 kgm/s

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) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note tha

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(a) 0.0021 s, 2926.5 rad/s

The frequency of the B note is

$f= 466 Hz$

The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{466 Hz}=0.0021 s$

The angular frequency instead is given by

$\omega = 2\pi f$

And substituting

f = 466 Hz

We find

$\omega = 2\pi (466 Hz)=2926.5 rad/s$

(b) 20 Hz, 125.6 rad/s

In this case, the period of the sound wave is

T = 50.0 ms = 0.050 s

So the frequency is equal to the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{0.050 s}=20 Hz$

While the angular frequency is given by:

$\omega = 2\pi f = 2 \pi (20 Hz)=125.6 rad/s$

(c) $4.30\cdot 10^{14} Hz, 7.48\cdot 1^{14} Hz, 2.33\cdot 10^{-15} s, 1.34\cdot 10^{-15}s$

The minimum angular frequency of the light wave is

$\omega_1 = 2.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{2.7\cdot 10^{15}rad/s}{2\pi}=4.30\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{4.30\cdot 10^{14}Hz}=2.33\cdot 10^{-15}s$

The maximum angular frequency of the light wave is

$\omega_2 = 4.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{4.7\cdot 10^{15}rad/s}{2\pi}=7.48\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{7.48\cdot 10^{14}Hz}=1.34\cdot 10^{-15}s$

(d) $2.0\cdot 10^{-7}s, 3.14\cdot 10^{7} rad/s$

In this case, the frequency is

$f=5.0 MHz = 5.0 \cdot 10^6 Hz$

So the period in this case is

$T=\frac{1}{f}=\frac{1}{5.0\cdot 10^6 Hz}=2.0 \cdot 10^{-7} s$

While the angular frequency is given by

$\omega = 2\pi f=2 \pi (5.0\cdot 10^{6}Hz)=3.14\cdot 10^{7} rad/s$

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3 years ago

A 61 kg skater is traveling at 2.5 m/s while carrying a 4.0 kg bowling ball. After he throws the bowling ball forward at twice t

gregori [183]

The final velocity of the skater is 2.34 m/s forward

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the system before and after the ball is thrown must be conserved, in absence of external forces.

Before the ball is thrown, the total momentum is:

$p_i = (M+m)u$

where

M = 61 kg is the mass of the skater

m = 4.0 kg is the mass of the ball

u = 2.5 m/s (forward) is the combined velocity of the skater and the ball

After, the ball is thrown at twice the velocity, so the final total momentum is

$p_f = MV+mv$

where

V is the final velocity of the skater

v = 2(2.5) = 5.0 is the final velocity of the ball

Since the total momentum must be conserved, we can write

$p_i = p_f\\(M+m)u = MV+mv\\V=\frac{(M+m)u-mv}{M}=\frac{(61+4.0)(2.5)-(4.0)(5.0)}{61}=2.34 m/s$

So, the skater is moving at 2.34 m/s (forward) after the shot.

Learn more about momentum:

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